Project Euler - 001

icymirror

最近沒有那么忙，開始學(xué)習(xí)golang，有空就拿Project Euler上的題目來練手吧。

Problem 1.
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.

問題1：
如果我們把10以內(nèi)的3或5的倍數(shù)列出來，我們會得到：3, 5, 6和9。這些數(shù)之和是23。
現(xiàn)在，找出1000以內(nèi)的3或5的倍數(shù)的數(shù)字之和。

method 1

1. package main
   
2. 
   
3. import (
   
4.         "fmt"
   
5. )
   
6. 
   
7. func Problem001_basic(scope, factor1, factor2 int) int {
   
8.         sum := 0
   
9. 
   
10.         for index := 0; index <= scope; index += factor1 {
    
11.                 sum = sum + index
    
12.         }
    
13.        
    
14.         for index := 0; index <= scope; index += factor2 {
    
15.                 sum = sum + index
    
16.         }
    
17.        
    
18.         for index := 0; index <= scope; index += factor1 * factor2 {
    
19.                 sum = sum - index
    
20.         }
    
21.        
    
22.         return sum
    
23. }
    
24. 
    
25. func Problem001_improve(scope, factor1, factor2 int) int {
    
26.         sum := (factor1 + scope / factor1 * factor1) * (scope / factor1) / 2
    
27.         sum = sum + (factor2 + scope / factor2 * factor2) * (scope / factor2) / 2
    
28.         factor3 := factor1 * factor2
    
29.         sum = sum -  (factor3 + scope / factor3 * factor3) * (scope / factor3) / 2
    
30.         return sum
    
31. }
    
32. 
    
33. func main() {
    
34.         fmt.Println("Problem 001 result: ", Problem001_basic(1000, 3, 5))
    
35.         fmt.Println("Problem 001 result: ", Problem001_improve(1000, 3, 5))
    
36. }

復(fù)制代碼

ba_du_co

拿來練練手。
233168

1. #!perl6
   
2. say [+] (^1000).grep: * %% (3|5);
   

復(fù)制代碼

zsszss0000

結(jié)果是 234168吧  ！回復(fù) 2# ba_du_co